Question

2
$\sqrt{3 + 4 \sqrt{3} }$
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Answer 1

$\huge \boxed{\mathsf{\underline{radica \ell ~dup\ell o}}}$

$\boxed{\boxed{\boxed{\boxed{\mathsf{\sqrt{a\pm\sqrt{b}}=\sqrt{\dfrac{a+c}{2}}\pm\sqrt{\dfrac{a-c}{2}}}}}}}$

Onde

$\huge\boxed{\mathsf{c=\sqrt{a^2-b}}}$

$\dotfill$

$\mathsf{4\sqrt{3}=\sqrt{4^2\cdot3}=\sqrt{48}}$

$\mathsf{\sqrt{3+4\sqrt{3}}=\sqrt{3+\sqrt{48}}}\\\mathsf{c=\sqrt{3^2-48}=\sqrt{-39}\not\in\mathbb{R}}$

Não possível transformar em radical duplo.